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Math Help - Series dealing with exponationals

  1. #1
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    Series dealing with exponationals

    I don't see how

    1. \sum_{n=1}^\infty \frac{1}{e^n} + \frac{1}{n(n+1)} = \frac{e}{e-1}

    2. \sum_{n=1}^\infty e^\frac{1}{n} - e^\frac{1}{n+1} = e-1

    In the first case I know that as n\rightarrow\infty that \frac{1}{e^n} approaches 0 and therefore it converges, and I know that \frac{1}{n(n+1)} = 1 (from a previous problem). But how do you know what \frac{1}{e^n} sums up to?.

    And in the second case I have no clue about approaching the problem, but I would think that it's the same type of approach as the first.
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  2. #2
    Senior Member Peritus's Avatar
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    <br />
\sum\limits_{n = 1}^\infty  {\left( {\frac{1}<br />
{e}} \right)^n }

    is just an infinite geometric series, and we know that it converges since 1/e < 1 thus:


    \sum\limits_{n = 1}^\infty  {\left( {\frac{1}<br />
{e}} \right)^n }  = \frac{{a_1 }}<br />
{{1 - q}} = \frac{{\frac{1}<br />
{e}}}<br />
{{1 - \frac{1}<br />
{e}}} = \frac{1}<br />
{{e - 1}}

    the second term can be rewritten using partial fraction decomposition as:

    <br />
\sum\limits_{n = 1}^\infty  {\frac{1}<br />
{{n\left( {n + 1} \right)}} = } \sum\limits_{n = 1}^\infty  {\frac{1}<br />
{n}}  - \frac{1}<br />
{{n + 1}}

    <br />
\sum\limits_{n = 1}^\infty  {\frac{1}<br />
{n}}  - \frac{1}<br />
{{n + 1}} = \left( {1 - \frac{1}<br />
{2}} \right) + \left( {\frac{1}<br />
{2} - \frac{1}<br />
{3}} \right) + \left( {\frac{1}<br />
{3} - \frac{1}<br />
{4}} \right) +  \cdots  = 1<br />

    since both individual terms converge to 0.
    so:

    <br />
\sum\limits_{n = 1}^\infty  {\frac{1}<br />
{{e^n }}}  + \sum\limits_{n = 1}^\infty  {\frac{1}<br />
{{n\left( {n + 1} \right)}} = \frac{1}<br />
{{e - 1}} + 1 = \frac{e}<br />
{{e - 1}}} <br />

    <br />
\sum\limits_{n = 1}^\infty  {e^{\frac{1}<br />
{n}}  - e^{\frac{1}<br />
{{n + 1}}}  = \left( {e - e^{\frac{1}<br />
{2}} } \right) + \left( {e^{\frac{1}<br />
{2}}  - e^{\frac{1}<br />
{3}} } \right) +  \cdots } <br />

    now let's look at the sum of the first N elements and then take the limit as N approaches infinity:

    <br />
\mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = 1}^N {e^{\frac{1}<br />
{n}}  - e^{\frac{1}<br />
{{n + 1}}}  = \left( {e - e^{\frac{1}<br />
{2}} } \right) + \left( {e^{\frac{1}<br />
{2}}  - e^{\frac{1}<br />
{3}} } \right) +  \cdots } <br />
<br />
 + \left( {e^{\frac{1}<br />
{{N - 1}}}  - e^{\frac{1}<br />
{N}} } \right) + \left( {e^{\frac{1}<br />
{N}}  - e^{\frac{1}<br />
{{N + 1}}} } \right) = e - 1<br />
    Last edited by Peritus; February 18th 2008 at 12:49 AM.
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